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    One Way ANOVA Table for N = 18, k = 3 & α = 5%

    One & Two Way ANOVA Calculator

    One way ANOVA table or summary to estimate the test of hypothesis (H0) at 5% or 0.05 significance level for sample means by using F-distribution table


    One-Way ANOVA Classification Table
    Sources of VariationdfSSMSSF-Ratio
    between Treatments210.411110.4111/2

    = 5.2056
    5.2056/3.2467

    = 1.6034
    Error1548.748.7/15

    = 3.2467
     
    Total17   

    Work with Steps for N = 18, k = 3 & α = 5%

    The below is the example work with steps shows how to generate the one way ANOVA table for the different means of treatments to find the F test statistic to test hypothesis (H0) at a 5% or 0.05 significance level for three sample means x1, x2 & x3.

    Workout :

    step 1 Address the formula input parameters and values
    x1 = 10, 12, 13, 11, 10, 13 & 15
    x2 = 9, 11, 13, 9 & 11
    x3 = 11, 15, 13, 12, 14 & 10

    step 2 Form the below table to carry out the analysis of variance

    Totalsquare
    x110121311101315 ∑x1 = 84 (∑x1)2 = 7056
    x291113911 ∑x2 = 53 (∑x2)2 = 2809
    x3111513121410 ∑x3 = 75 (∑x3)2 = 5625
    G = 212 ∑Ti2 = 15490

    Individual Squares :
    Total
    x12100144169121100169225 ∑x12 = 1028
    x228112116981121 ∑x22 = 573
    x32121225169144196100 ∑x32 = 955
    ∑xij2 = 2556

    step 3 Find Correction Factor (CF)
    N = total number of elements in all three sample dataset.
    N = 7 + 5 + 6
    N = 18
    Correction factor (CF) = G2N
    CF = 212218
    = 44944/18

    CF = 2496.8889

    step 4 Find the sum of squares of all individual items (∑xij) and then total sum of squares (TSS)
    Total sum of squares (TSS) = ∑∑xij2 - CF
    = 2556 - 2496.8889
    TSS = 59.1111

    step 5 Find the sum of squares between the treatments (SST)

    SST = k i = 1 Ti2ni - CF
    = (∑x1)2n1+(∑x2)2n2+(∑x3)2n3 - CF
    = 7056/7+2809/5+5625/6 - 2496.8889
    = (1008 + 561.8 + 937.5) - 2496.8889
    = 2507.3 - 2496.8889
    = 10.4111

    step 6 Find sum of squares due to the error (SSE)
    SSE = TSS - SST
    = 59.1111 - 10.4111
    = 48.7

    step 7 The above sum of squares together with their respective degrees of freedom and mean sum of squares will be summarized in the following table.

    Sources of VariationdfSSMSSF-ratio
    between Treatmentsk - 1SSTSST/k-1

    = MST
    MST/MSE

    = F T
    ErrorN - kSSESSE/N-k

    = MSE
     
    TotalN - 1   

    One-Way ANOVA Classification Table
    Sources of Variation   df   SSMSSF-Ratio
    between Treatment 3-1 = 210.411110.4111/2

    = 5.2056
    5.2056/3.2467

    = 1.6034
    Error 18-3 = 1548.748.7/15

    = 3.2467
     
    Total 18-1 = 17   

    step 8 Find Critical value of Fe or Table value of F
    The critical value of Fe from the F-distribution table for the degrees of freedom (k - 1, N - k) at 5% significance level.
    For treatments, the critical value Fe(2,15) from F-distribution table at 5% of significance level is 3.6823

    Inference
    There is no significance difference between treatments, since the calculated value of F0 = 1.6034 is smaller than the table value of Fe = 3.6823. Therefore the null hypothesis H0 is accepted.

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